1. **State the problem:** We have three charges along the x-axis: $q_1$ at $x = -2.00$ cm, $q_3 = 5$ nC at the origin, and $q_2 = -3$ nC at $x = 4$ cm. We want to find the charge $q_1$ such that the net force on $q_3$ is zero. 2. **Relevant formula:** The electrostatic force between two point charges is given by Coulomb's law: $$F = k \frac{|q_a q_b|}{r^2}$$ where $k = 8.99 \times 10^9$ N m$^2$/C$^2$, $q_a$ and $q_b$ are the charges, and $r$ is the distance between them. 3. **Set up forces on $q_3$:** - Force on $q_3$ due to $q_1$ is $F_{31} = k \frac{|q_3 q_1|}{(0.02)^2}$ (distance 2 cm = 0.02 m). - Force on $q_3$ due to $q_2$ is $F_{32} = k \frac{|q_3 q_2|}{(0.04)^2}$ (distance 4 cm = 0.04 m). 4. **Direction of forces:** - $q_3$ is positive (5 nC). - $q_2$ is negative (-3 nC), so force on $q_3$ due to $q_2$ is attractive, pulling $q_3$ toward $q_2$ (positive x direction). - To have net force zero, force from $q_1$ must balance this, so force from $q_1$ must be equal in magnitude and opposite in direction (negative x direction). 5. **Write force balance equation:** $$F_{31} = F_{32}$$ $$k \frac{|q_3 q_1|}{(0.02)^2} = k \frac{|q_3 q_2|}{(0.04)^2}$$ 6. **Simplify and solve for $q_1$:** Cancel $k$ and $q_3$ (both positive): $$\frac{|q_1|}{(0.02)^2} = \frac{|q_2|}{(0.04)^2}$$ Calculate denominators: $$(0.02)^2 = 0.0004, \quad (0.04)^2 = 0.0016$$ So: $$\frac{|q_1|}{0.0004} = \frac{3 \times 10^{-9}}{0.0016}$$ Multiply both sides by $0.0004$: $$|q_1| = 3 \times 10^{-9} \times \frac{0.0004}{0.0016} = 3 \times 10^{-9} \times 0.25 = 0.75 \times 10^{-9} = 7.5 \times 10^{-10}$$ 7. **Determine sign of $q_1$:** Since force from $q_1$ must be opposite to force from $q_2$ and $q_2$ is negative, $q_1$ must be positive to repel $q_3$ (which is positive) to the left. **Final answer:** $$q_1 = +7.5 \times 10^{-10} \text{ C} = +0.75 \text{ nC}$$